Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
inline int sgn(double x) { return x < -1e-8 ? -1 : (x > 1e-8 ? 1 : 0); }
int n, q;
double sy, a, b;
pair<double, double> itv[MAXN];
double pre[MAXN];
double calc(double l, double r) {
  int i = lower_bound(itv + 1, itv + n + 1, make_pair(l, l),
                      [](const pair<double, double>& a,
                         const pair<double, double>& b) -> bool {
                        return sgn(a.first - b.first) < 0;
                      }) -
          itv - 1;
  int j = upper_bound(itv + 1, itv + n + 1, make_pair(r, r),
                      [](const pair<double, double>& a,
                         const pair<double, double>& b) -> bool {
                        return sgn(a.second - b.second) < 0;
                      }) -
          itv;
  double ret = (i < j ? pre[j - 1] - pre[i] : 0);
  if (i && sgn(itv[i].second - l) > 0) ret += min(r, itv[i].second) - l;
  if (j > i && j <= n && sgn(itv[j].first - r) < 0)
    ret += r - max(l, itv[j].first);
  return ret;
}
int main() {
  scanf("%lf%lf%lf", &sy, &a, &b);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lf%lf", &itv[i].first, &itv[i].second);
  sort(itv + 1, itv + n + 1);
  for (int i = 1; i <= n; i++)
    pre[i] = pre[i - 1] + itv[i].second - itv[i].first;
  scanf("%d", &q);
  while (q--) {
    double x, y;
    scanf("%lf%lf", &x, &y);
    double l = (y * a - x * sy) / (y - sy);
    double r = (y * b - x * sy) / (y - sy);
    double len = calc(l, r);
    double ans = len * (b - a) / (r - l);
    printf("%.12lf\n", ans);
  }
  return 0;
}